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# Trig identities

## by Simon Borgert on Jun 14, 2011

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Basic Trig Identities

Basic Trig Identities

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## Trig identitiesPresentation Transcript

• Trig Identities Tuesday 14th June 2011
• y P(x, y) 1 x
• y P(x, y) 1 x
• y P(x, y) 1 xAngles are always taken in an anticlockwise direction from the positive x axis
• y P(x, y) 1 θ xAngles are always taken in an anticlockwise direction from the positive x axis
• y P(x, y) 1 θ x
• y P(x, y) 1 θ xConsider the coordinates of P
• y P(x, y) 1 θ x xConsider the coordinates of P
• y y P(x, y) 1 θ x xConsider the coordinates of P
• y P(x, y) 1 θ x
• y P(x, y) 1 θ xThis creates a right angled triangle with an angle θ
• y P(x, y) 1 θ xThis creates a right angled triangle with an angle θ
• y P(x, y) 1 θ x xThis creates a right angled triangle with an angle θ
• y P(x, y) 1 y θ x xThis creates a right angled triangle with an angle θ
• y P(x, y) 1 y θ x x
• y P(x, y) 1 y θ x xUsing our standard trigonometric ratios
• y y tan θ = x P(x, y) 1 y θ x xUsing our standard trigonometric ratios
• y y tan θ = x P(x, y) y sin θ = = y 1 1 y θ x xUsing our standard trigonometric ratios
• y y tan θ = x P(x, y) y sin θ = = y 1 1 y x cosθ = = x θ 1 x xUsing our standard trigonometric ratios
• y 1st Quadrant tan θ = y x P(x, y) y sin θ = = y 1 1 y x cosθ = = x θ 1 x x
• y 1st Quadrant tan θ = y x P(x, y) y sin θ = = y 1 1 y x cosθ = = x θ 1 x x(By Pythagoras):
• y 1st Quadrant tan θ = y x P(x, y) y sin θ = = y 1 1 y x cosθ = = x θ 1 x x 2 2 x + y =1(By Pythagoras):
• y 1st Quadrant tan θ = y x P(x, y) y sin θ = = y 1 1 y x cosθ = = x θ 1 x x 2 2 x + y =1(By Pythagoras): 2 2 ∴sin θ + cos θ = 1
• VERY important Trig Identity
• VERY important Trig Identity 2 2 sin θ + cos θ = 1
• or rearranging
• or rearranging 2 2sin θ + cos θ = 1
• or rearranging 2 2sin θ + cos θ = 1 2 2sin θ = 1 − cos θ
• or rearranging 2 2sin θ + cos θ = 1 2 2sin θ = 1 − cos θ 2 2cos θ = 1 − sin θ
• or dividing both sides
• or dividing both sides 2 2 sin θ + cos θ = 1
• or dividing both sides 2 2 sin θ + cos θ = 1 2 2sin θ cos θ 1 2 + 2 = 2sin θ sin θ sin θ
• or dividing both sides
• or dividing both sides 2 2 sin θ + cos θ = 1
• or dividing both sides 2 2 sin θ + cos θ = 1 2 2sin θ cos θ 1 2 + 2 = 2sin θ sin θ sin θ
• or dividing both sides 2 2 sin θ + cos θ = 1 2 2 sin θ cos θ 1 2 + 2 = 2 sin θ sin θ sin θ 2 2∴1 + cot θ = cosec θ
• or dividing both sides
• or dividing both sides 2 2 sin θ + cos θ = 1
• or dividing both sides 2 2 sin θ + cos θ = 1 2 2sin θ cos θ 1 2 + 2 = 2sin θ sin θ sin θ
• or dividing both sides
• or dividing both sides 2 2 sin θ + cos θ = 1
• or dividing both sides 2 2 sin θ + cos θ = 1 2 2sin θ cos θ 1 2 + 2 = 2cos θ cos θ cos θ
• or dividing both sides
• or dividing both sides 2 2 sin θ + cos θ = 1
• or dividing both sides 2 2 sin θ + cos θ = 1 2 2sin θ cos θ 1 2 + 2 = 2cos θ cos θ cos θ
• or dividing both sides 2 2 sin θ + cos θ = 1 2 2 sin θ cos θ 1 2 + 2 = 2 cos θ cos θ cos θ 2 2∴ tan θ + 1 = sec θ
• or dividing both sides
• or dividing both sides 2 2 sin θ + cos θ = 1
• or dividing both sides 2 2 sin θ + cos θ = 1 2 2sin θ cos θ 1 2 + 2 = 2cos θ cos θ cos θ
• or dividing both sides 2 2 sin θ + cos θ = 1 2 2 sin θ cos θ 1 2 + 2 = 2 cos θ cos θ cos θ 2 2∴ tan θ + 1 = sec θ
• or dividing both sides 2 2 sin θ + cos θ = 1 2 2 sin θ cos θ 1 2 + 2 = 2 cos θ cos θ cos θ 2 2∴ tan θ + 1 = sec θ 2 2 ∴ tan θ = sec θ − 1
• or dividing both sides
• or dividing both sides 2 2 sin θ + cos θ = 1
• or dividing both sides 2 2 sin θ + cos θ = 1 2 2sin θ cos θ 1 2 + 2 = 2cos θ cos θ cos θ
• or dividing both sides
• or dividing both sides 2 2 sin θ + cos θ = 1
• or dividing both sides 2 2 sin θ + cos θ = 1 2 2sin θ cos θ 1 2 + 2 = 2cos θ cos θ cos θ
• or dividing both sides 2 2 sin θ + cos θ = 1 2 2 sin θ cos θ 1 2 + 2 = 2 cos θ cos θ cos θ 2 2∴ tan θ + 1 = sec θ
• or dividing both sides
• or dividing both sides 2 2 sin θ + cos θ = 1
• or dividing both sides 2 2 sin θ + cos θ = 1 2 2sin θ cos θ 1 2 + 2 = 2cos θ cos θ cos θ
• or dividing both sides 2 2 sin θ + cos θ = 1 2 2 sin θ cos θ 1 2 + 2 = 2 cos θ cos θ cos θ 2 2∴ tan θ + 1 = sec θ
• or dividing both sides 2 2 sin θ + cos θ = 1 2 2 sin θ cos θ 1 2 + 2 = 2 cos θ cos θ cos θ 2 2∴ tan θ + 1 = sec θ 2 2 ∴ tan θ = sec θ − 1
• But easier to remember how to get them!
• 2 2 sin θ + cos θ = 1But easier to remember how to get them!
• 2 2 sin θ + cos θ = 1 2 2cosec θ = 1 + cot θBut easier to remember how to get them!
• 2 2 sin θ + cos θ = 1 2 2cosec θ = 1 + cot θ 2 2 cot θ = cosec θ − 1But easier to remember how to get them!
• 2 2 sin θ + cos θ = 1 2 2cosec θ = 1 + cot θ 2 2 cot θ = cosec θ − 1 2 2 sec θ = tan θ + 1But easier to remember how to get them!
• 2 2 sin θ + cos θ = 1 2 2cosec θ = 1 + cot θ 2 2 cot θ = cosec θ − 1 2 2 sec θ = tan θ + 1 2 2∴ tan θ = sec θ − 1But easier to remember how to get them!
• 2 2 sin θ + cos θ = 1 2 2 sin θ = 1 − cos θ 2 2cosec θ = 1 + cot θ 2 2 cot θ = cosec θ − 1 2 2 sec θ = tan θ + 1 2 2∴ tan θ = sec θ − 1But easier to remember how to get them!
• 2 2 sin θ + cos θ = 1 2 2 sin θ = 1 − cos θ 2 2 cos θ = 1 − sin θ 2 2cosec θ = 1 + cot θ 2 2 cot θ = cosec θ − 1 2 2 sec θ = tan θ + 1 2 2∴ tan θ = sec θ − 1But easier to remember how to get them!
• ExampleProve 2 1 − sin θ 2 2 2 = cos θ sin θ + cos θ
• ExampleProve 2 1 − sin θ 2 2 2 = cos θ sin θ + cos θ Only work on one side at a time!
• ExampleProve 2 1 − sin θ 2 2 2 = cos θ sin θ + cos θ 2 1 − sin θ LHS = 2 2 sin θ + cos θ
• ExampleProve 2 1 − sin θ 2 2 2 = cos θ sin θ + cos θ 2 1 − sin θ LHS = 2 2 sin θ + cos θ 2 cos θ = 1
• ExampleProve 2 1 − sin θ 2 2 2 = cos θ sin θ + cos θ 2 1 − sin θ LHS = 2 2 sin θ + cos θ 2 cos θ = 1 2 = cos θ
• Example cosθProve − tan θ = sec θ 1 − sin θ
• Example cosθProve − tan θ = sec θ 1 − sin θ cosθ sin θ LHS = − 1 − sin θ cosθ
• Example cosθProve − tan θ = sec θ 1 − sin θ cosθ sin θ LHS = − 1 − sin θ cosθ (cosθ )cosθ sin θ (1 − sin θ ) Common = − (1 − sin θ )cosθ cosθ (1 − sin θ ) Denominator
• Example cosθProve − tan θ = sec θ 1 − sin θ cosθ sin θ LHS = − 1 − sin θ cosθ (cosθ )cosθ sin θ (1 − sin θ ) = − (1 − sin θ )cosθ cosθ (1 − sin θ ) 2 2 cos θ − sin θ + sin θ = (1 − sin θ )cosθ
• Example cosθProve − tan θ = sec θ 1 − sin θ cosθ sin θ LHS = − 1 − sin θ cosθ (cosθ )cosθ sin θ (1 − sin θ ) = − (1 − sin θ )cosθ cosθ (1 − sin θ ) 2 2 cos θ − sin θ + sin θ = (1 − sin θ )cosθ 1 − sin θ = (1 − sin θ )cosθ
• Example cosθProve − tan θ = sec θ 1 − sin θ cosθ sin θ LHS = − 1 − sin θ cosθ (cosθ )cosθ sin θ (1 − sin θ ) = − (1 − sin θ )cosθ cosθ (1 − sin θ ) 2 2 cos θ − sin θ + sin θ = (1 − sin θ )cosθ 1 − sin θ 1 = = (1 − sin θ )cosθ cosθ
• Example cosθProve − tan θ = sec θ 1 − sin θ cosθ sin θ LHS = − 1 − sin θ cosθ (cosθ )cosθ sin θ (1 − sin θ ) Common = − (1 − sin θ )cosθ cosθ (1 − sin θ ) Denominator 2 2 cos θ − sin θ + sin θ = (1 − sin θ )cosθ 1 − sin θ 1 = = = sec θ (1 − sin θ )cosθ cosθ
• Example 3 o o If sin θ = and 90 ≤ θ ≤ 180 find cosθ and tanθ 5
• Example 3 o o If sin θ = and 90 ≤ θ ≤ 180 find cosθ and tanθ 5The restrictions tell us which quadrant
• Example 3 o o If sin θ = and 90 ≤ θ ≤ 180 find cosθ and tanθ 5The restrictions tell us which quadrant 2nd!
• Example 3 o o If sin θ = and 90 ≤ θ ≤ 180 find cosθ and tanθ 5The restrictions tell us which quadrant 2nd!Think about the triangle involved 3 O since sin θ = = 5 H θ
• Example 3 o o If sin θ = and 90 ≤ θ ≤ 180 find cosθ and tanθ 5The restrictions tell us which quadrant 2nd!Think about the triangle involved 3 O since sin θ = = 5 5 H 3 θ
• Example 3 o o If sin θ = and 90 ≤ θ ≤ 180 find cosθ and tanθ 5The restrictions tell us which quadrant 2nd!Think about the triangle involved 5 By Pythagoras! 3 θ 4
• Example 3 o o If sin θ = and 90 ≤ θ ≤ 180 find cosθ and tanθ 5The restrictions tell us which quadrant 2nd!Think about the triangle involved −3 tan θ = 5 4 3 θ 4
• Example 3 o o If sin θ = and 90 ≤ θ ≤ 180 find cosθ and tanθ 5The restrictions tell us which quadrant 2nd!Think about the triangle involved −3 5 tan θ = 4 3 θ −4 cosθ = 4 5
• Now do Ex 8.13